modelling with differential equations 8d · modelling with differential equations 8d 1 the system...

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Modelling with differential equations 8D

1 The system of equations is

Rearranging equation (1) and differentiating with respect to t gives:

Substituting into equation (2) gives:

Solving the auxiliary equation

So Then differentiating with respect to t and substituting in equation (3) gives:

Using the initial conditions at , and gives (4)

(5) Substituting equation (4) into equation (5)

Substituting for A and B gives the particular solutions

d ( )dd ( )d

x x yty x yt

= +

= -

1

2

2

2

d ( )d

d d dd d d

xy xt

y x xt t t

= -

= -

3

d2xdt2 − dx

dt= x − dx

dt+ x

d2xdt2 − 2x = 0

m2 − 2 = 0⇒ m = ± 2

x = Ae 2t + Be− 2t

y = dxdt

− x = 2Ae 2t − 2Be− 2t − Ae 2t − Be− 2t

y = A( 2 −1)e 2t − B( 2 +1)e− 2t

t = 0 x = 1 y = 2 A+ B = 1⇒ A = 1− B

A( 2 −1)− B( 2 +1) = 2

(1− B)( 2 −1)− B( 2 +1) = 2

⇒ 2 −1− 2B + B − 2B − B = 2

⇒ B = 2 − 3

2 2= 2− 3 2

4

So A = 1− 2− 3 24

= 2+ 3 24

x = 2+ 3 24

e 2t + 2− 3 24

e− 2t = 14

(2+ 3 2)e 2t + 14

(2− 3 2)e− 2t

y = 2+ 3 24

( 2 −1)e 2t − 2− 3 24

( 2 +1)e− 2t

= 4− 24

e 2t + 4+ 24

e− 2t = 14

(4− 2)e 2t + 14

(4+ 2)e− 2t


2 a The system of equations is




So Then differentiating with respect to t

So the general solutions are and

Note that the problem can also be solved by rearranging equation (1) to obtain

Then differentiating with respect to t and substituting in equation (2) and following a similar methodology gives the mathematically equivalent results:

and

b At , so


d 5 ( )dd 3 ( )d

x x yty y xt

= +

= - -

1

2

x = −3y − dydt

dxdt

= −3dydt

− d2 ydt2

−3dydt

− d2 ydt2 = −3y − dy

dt+5y

d2 ydt2 + 2 dy

dt+ 2y = 0

m2 + 2m+ 2 = 0

m = −2 ± 4−82

= −1± i

y = e− t ( Acos t + Bsin t)

d e ( cos sin ) e ( sin cos )dSubstituting in equation ( ) gives:

e ( cos sin ) e ( sin cos ) 3(e ( cos sin ))e ( cos sin sin cos 3 cos 3 sin )

e ((2 )cos (2

t t

t t t

t

t

y A t B t A t B tt

A t B t A t B t A t B t xx A t B t A t B t A t B tx A B t B

- -

- - -

-

-

= - + + - +

- + + - + = - + -

Þ = + + - - -

Þ = - + +

2

)sin )A t-

x = −e− t ((2A+ B)cos t + (2B − A)sin t) y = e− t ( Acos t + Bsin t)

5y = − dx

dt+ x

x = e− t ( Acos t + Bsin t) y = 1

5e− t ((B − 2A)cos t − ( A+ 2B)sin t)

t = 0 x = 1 and y = 2

From the equation for x: − (2A+ B) = 1⇒ B = −1− 2AFrom the equation for y: A = 2⇒ B = −5

x = e− t (cos t +12sin t)

y = e− t (2cos t −5sin t)


3 The system of equations is




So

Then differentiating with respect to t

Substituting into equation (2) gives

Using the initial conditions at , and gives:


d 2 3 2 ( )dd 1 ( )d

x x yty x yt

= - -

= + -

1

2

x = dydt

− y +1

dxdt

= d2 ydt2 − dy

dt

d2 ydt2 − dy

dt= 2 dy

dt− 2y + 2− 3y − 2

d2 ydt2 − 3dy

dt+5y = 0

m2 − 3m+5= 0⇒ m = 3± 9− 20

2= 3

2± 11

2

y = e

3t2 Acos

112

t + Bsin112

t⎛

⎝⎜

⎞

⎠⎟

dydt

= 32

e3t2 Acos

112

t + Bsin112

t⎛

⎝⎜

⎞

⎠⎟ + e

3t2 − 11

2Asin

112

t + 112

Bcos112

t⎛

⎝⎜

⎞

⎠⎟

= e3t2 3

2A+ 11

2B

⎛

⎝⎜

⎞

⎠⎟ cos

112

t + 32

B − 112

A⎛

⎝⎜

⎞

⎠⎟ sin

112

t⎛

⎝⎜

⎞

⎠⎟

e3t2 3

2A+ 11

2B

⎛

⎝⎜

⎞

⎠⎟ cos

112

t + 32

B − 112

A⎛

⎝⎜

⎞

⎠⎟ sin

112

t⎛

⎝⎜

⎞

⎠⎟ = x + e

3t2 Acos

112

t + Bsin112

t⎛

⎝⎜

⎞

⎠⎟ −1

⇒ x = e3t2 1

2A+ 11

2B

⎛

⎝⎜

⎞

⎠⎟ cos

112

t + 12

B − 112

A⎛

⎝⎜

⎞

⎠⎟ sin

112

t⎛

⎝⎜

⎞

⎠⎟ +1

t = 0 x = 0 y = 1

12

A+ 112

B +1= 0⇒ A = − 11B − 2 and A = 1, so B = − 3

11

x = e3t2 1

2− 3 11

2 11

⎛

⎝⎜

⎞

⎠⎟ cos

112

t + − 3

2 11− 11

2

⎛

⎝⎜

⎞

⎠⎟ sin

112

t⎛

⎝⎜

⎞

⎠⎟ +1= e

3t2 −cos

112

t − 7

11sin

112

t⎛

⎝⎜

⎞

⎠⎟ +1

y = e

3t2 cos

112

t − 3

11sin

112

t⎛

⎝⎜

⎞

⎠⎟





Dividing through by 5 gives:

b To find x, first solve the auxiliary equation

So c To find y, first differentiate the equation for x from part b

Then substituting into equation (3)

d 0.2 0.2 ( )dd 0.5 0.4 ( )d

x x yty x yt

= +

= - +

1

2

2

2

d5 ( )d

d d d5d d d

xy xt

y x xt t t

= -

= -

3

5d2xdt2 − dx

dt= −0.5x + 2 dx

dt− 0.4x

5d2xdt2 − 3dx

dt+ 0.9x = 0

d2xdt2 − 0.6 dx

dt+ 0.18x = 0

m2 − 0.6m+ 0.18 = 0

m = 0.6 ± 0.36− 0.722

= 0.6 ± −0.362

= 0.3± 0.3i

x = e0.3t ( Acos0.3t + Bsin0.3t)

dxdt

= e0.3t (−0.3Asin0.3t + 0.3Bcos0.3t)+ 0.3e0.3t ( Acos0.3t + Bsin0.3t)

= 0.3e0.3t (B − A)sin0.3t + 0.3e0.3t ( A+ B)cos0.3t

y = 5dxdt

− x = 5 0.3e0.3t (B − A)sin0.3t + 0.3e0.3t ( A+ B)cos0.3t( )− e0.3t ( Acos0.3t + Bsin0.3t)

= e0.3t (1.5A+1.5B − A)cos0.3t + (1.5B −1.5A− B)sin0.3t( )= 0.1e0.3t (5A+15B)cos0.3t + (5B −15A)sin0.3t( )


4 d At , there were 3 sand foxes and 111 meerkats on the island, i.e. . So:

So need to find t for which y = 0

So, since the first measurement was taken in 2012, the meerkats will die out some time in 2018. e In the year when the meerkats die out, , so

The number of foxes has to be a natural number, so round to 441 foxes. f The model seems reasonable for the first few years, where both the number of meerkats and the

foxes are sensible. When after 6 years (see part d) the meerkats die out, the model becomes unsuitable. Considering larger values of t shows that the number of meerkats becomes negative. For instance, consider t = 10, we have

which is not a feasible number of animals. Similarly, for the number of foxes, if we consider

t = 15, we get:

This behaviour can also be seen from the equations for x and y – as the coefficient of cos and sin

are very large, they will lead to large negative expressions inside the brackets. Since the exponential on front of the bracket is always positive, the expressions will periodically become negative, making this model unsuitable for modelling the number of animals in the long term.

t = 0 x = 3 and y = 111

e0 Acos0+ Bsin0 = 3⇒ A = 3

0.1e0(5A+15B)cos0 = 111⇒ 0.1(15+15B) = 111⇒111= 1.5+1.5B ⇒ B = 73

0.3

0.3

0.1 (1110cos(0.3 ) 320sin(0.3 ))0 0.1 (1110cos(0.3 ) 320sin(0.3 ))0 1110cos(0.3 ) 320sin(0.3 )

1110cos(0.3 ) 320sin(0.3 )1110 tan(0.3 )3200.3 1.8515...

6.17 years (3 s.f.)

t

t

y e t te t t

t tt t

t

tt

= += += += -

- =

==

t = 60.3

0.3 6.17

(3cos(0.3 ) 73sin(0.3 ))(3cos(0.3 6.17) 73sin(0.3 6.17))

441 (3 s.f.)

tx e t te ´

= += ´ + ´=

30.1e (1110cos3 320sin3) 2096.5y = + » -

x = e0.3×15 3cos(0.3×15)+ 73sin(0.3×15)( ) ≈ −6480.521




Substituting into equation (2)


So Then differentiating with respect to t


Using the initial conditions, at , . Hence:


b c As t gets large, becomes much larger than t, so So the amounts of both chemicals in

the tank tend to zero.

d 3 2 ( )dd 2 ( )d

x x yty x yt

= - +

= - +

1

2

2

2

d1.5 0.5 ( )d

d d d1.5 0.5d d d

xy xt

y x xt t t

= +

= +

3

1.5dxdt

+ 0.5d2xdt2 = −2x +1.5x + 0.5dx

dtd2xdt2 + 2 dx

dt+ x = 0

m2 + 2m+1= 0(m+1)(m+1) = 0

m = −1

x = ( A+ Bt)e− t

dxdt

= Be− t − ( A+ Bt)e− t = e− t (B − A− Bt)

y = 1.5( A+ Bt)e− t + 0.5e− t (B − A− Bt)= ( A+ 0.5B + Bt)e− t

t = 0 x = 1 and y = 2

( A+ B × 0)e0 = 1⇒ A = 1( A+ 0.5B + B × 0)e0 = 2⇒ B = 2

x = (1+ 2t)e− t = Pe− t where P(t) = 1+ 2t

y = (2+ 2t)e− t = Qe− t where Q(t) = 2+ 2t

At t = 2, x = 5e−2 = 0.677 litres (3 d.p.) and y = 6e−2 = 0.812 litres (3 d.p.)

et

te− t → 0.





This equation describes simple harmonic motion for y. b Solving the auxiliary equation for equation (3) So Then differentiating with respect to t


Using the initial conditions, at . So:


d 4 ( )dd 4 ( )d

x yty xt

= -

=

1

2

2

2

d0.25 ( )d

d d0.25d d

yxt

x yt t

=

=

3

2

2

2

2

d0.25 4d

d 16 ( )d

y yt

y yt

= -

= - 4

m2 +16 = 0⇒ m = ±4i

y = Acos4t + Bsin4t

dydt

= −4Asin4t + 4Bcos4t

x = Bcos4t − Asin4t

t = 0, x = 4 and y = 5

Bcos0− Asin0 = 4⇒ B = 4Acos0+ Bsin0 = 5⇒ A = 5

x = 4cos4t −5sin4ty = 5cos4t + 4sin4t





b Solving the auxiliary equation

So the complementary function is Try a constant for the particular integral,

Substituting into gives:

So Differentiating with respect to t


So the general solutions are c Since the exponential terms tend to zero in the expressions for both x and y, the total amount of

toxin in the blood will tend to and in the organs to .

d 0.03 0.01 50 ( )dd 0.01 0.03 ( )d

x x yty x yt

= - + +

= -

1

2

2

2

d100 3 5000 ( )d

d d d100 3d d d

xy xt

y x xt t t

= + -

= +

3

100 d2xdt2 + 3dx

dt= 0.01x − 3dx

dt− 0.09x +150

d2xdt2 + 0.06 dx

dt+ 0.0008x = 1.5

2 0.06 0.0008 0( 0.04)( 0.02) 0

0.04 or 0.02

m mm m

m

+ + =+ + =

= - -

x = Ae−0.04t + Be−0.02t

x = λ, so !x = !!x = 0

d2xdt2 + 0.006 dx

dt+ 0.0008x = 1.5

0.0008λ = 1.5⇒λ = 1875

x = Ae−0.04t + Be−0.02t +1875

dxdt

= −0.04Ae−0.04t − 0.02Be−0.02t

y = 100(−0.04Ae−0.04t − 0.02Be−0.02t )+ 3( Ae−0.04t + Be−0.02t +1875)−5000= Be−0.02t − Ae−0.04t + 625

x = Ae−0.04t + Be−0.02t +1875

y = Be−0.02t − Ae−0.04t + 625

xlim = 1875 mg ylim = 625 mg




Substitute into equation (2) gives:

b Solving the auxiliary equation

So the complementary function is Try a constant for the particular integral,

Substituting into gives:

So

Differentiating with respect to t


c Note that both x, y are dominated by the positive exponential term, so as t gets large, both

quantities will grow to infinity. Thus the model does not seem to be well suited to describe the amounts of nutrients.

d 2 1 ( )dd 4 2 ( )d

x x yty x yt

= - + +

= - + +

1

2

2

2

d 2 1 ( )d

d d d2d d d

xy xt

y x xt t t

= + -

= +

3

d2xdt2 + 2 dx

dt= 4x + dx

dt+ 2x −1+ 2

d2xdt2 + dx

dt− 6x = 1

2 6 0( 2)( 3) 0

2 or 3

m mm m

m

+ - =- + =

= -

x = Ae2t + Be−3t

x = λ, so !x = !!x = 0

d2xdt2 + dx

dt− 6x = 1

−6λ = 1⇒λ = − 1

6

x = Ae2t + Be−3t − 1

6

dxdt

= 2Ae2t − 3Be−3t

2 3 2 3

2 3

12 e 3 e 2 e e 16

44 e e3

t t t t

t t

y A B A B

A B

- -

-

æ ö= - + + - -ç ÷è ø

= - -


Challenge

The populations are stable if . So:

(1)

(2) Substituting for x from equation (2) into equation (1) gives

When y = 0, x = 0; when y = 4800, x = 0.02(4800) = 96 So ignoring the trivial case when both populations are 0, the solution is 96 owls and 4800 field mice.

dydt

= 2y − y2

6000− 60x

dxdt

= 0.02y − x

dxdt

= dydt

= 0

2y − y2

6000− 60x = 0

0.02y − x = 0⇒ x = 0.02y

2y − y2

6000− 60× 0.02y = 0

⇒ y2

6000− 0.8y = 0

⇒ y2 − 4800y = 0⇒ y( y − 4800) = 0So y = 0 or 4800

modelling with differential equations 8d · modelling with differential equations 8d 1 the system...

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